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Homework Help: Finding the Potential Energy Function

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Which of the following forces is conservative? (a) F = k(x,2y,3z), (b) F=k(y,x,0), and
    (c) F=k(-y,x,0). For those which are conservative, find the corresponding potential energy U, and verify by direct substitution that [itex]\vec{F} = - \nabla U[/itex]

    2. Relevant equations

    3. The attempt at a solution
    I have already ascertained which force fields are conservative, and now I am trying to find the potential energy associated with those force fields.

    [itex]\vec{F} = - \nabla U[/itex] implies that [itex]F_x \hat{i} + F_y \hat {j} + F_z \hat{k} = \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial x} \hat{j} + \frac{\partial U}{\partial x} \hat{k}[/itex].

    Using this fact, I set the components equal to equal other, and integrated each individual component. When I went to see if this was the correct answer, I found the answer was a linear combination of the three individual components. Why is that?
  2. jcsd
  3. Oct 13, 2013 #2


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    What did you get for U in case of (a)? Remember, U is a scalar function. It does not have components.

  4. Oct 14, 2013 #3
    \hat{}No, I understand that U is a scalar function, that it only has a scalar value associated with every point in space (only if that particular point is in the domain of the function U, of course). For part (a), I get [itex]U= - \frac{1}{2}kx^2 + U_0[/itex]. However, the actual answer is [itex]U=-\frac{1}{2}k(x^2 +2y^2 + 3z^2)[/itex]. Like mentioned in my first post, it appears that they took the individual vector components, integrated each one, and then added them together.

    Here is another way I thought of solving this problem:

    [itex]W =- \Delta U = \int _{\vec{r}_0}^\vec{r} F(\vec{r}')\cdot d \vec{r}'[/itex]. I can define the potential energy to be zero at [itex]\vec{r}_0 = \langle 0,0,0 \rangle[/itex], thus measuring changes in PE relative to the origin. [itex]\vec{r} \langle x',y',z' \rangle[/itex] will be some later position of the object whom the work is done by the force. Since the force is conservative, we need not concern ourselves with the path taken to get from r_0 to r; instead, we can just integrate along the straight line connecting the two points.

    [itex]\Delta U = - \int_{\vec{r}_0}^\vec{r} \langle F_x, F_y , F_z \rangle \cdot \langle dx , dy , dz \rangle = -\int_{\vec{r}_0}^\vec{r} (F_x dx + F_y dy + F_z dz)[/itex](1).

    The part am I having is with the limits of integration. If the integral could somehow become [itex]\Delta U = - \int_0^x F_x dx' - \int_0^y F_y dy' - \int_0^z F_z dz'[/itex] (2), then I would get the correct answer. The problem I am having difficulty with is going from (1) to (2). How do you justify such a step?
    Last edited: Oct 14, 2013
  5. Oct 14, 2013 #4
    I don't intend to be a bother, but could someone possibly help me?
  6. Oct 15, 2013 #5


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    As U is a potential function, the integration path is arbitrary. You can reach from 0 to (x,y,z) as going first from (0,0,0) to (x,0 0) then from (x,0.0) to (x,y,0) and then from (x,y,0) to (x,y,z).

  7. Oct 15, 2013 #6


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    You have that [itex]\dfrac{\partial u}{\partial x}= x[/itex], [itex]\dfrac{\partial u}{\partial y}= 2y[/itex], and [itex]\dfrac{\partial u}{\partial z}= 3z[/itex].

    Integrating the first equation, with respect to x, we have [itex]u= x^2/2+ f(y, z)[/itex]. (Since the partial derivative is with respect to the single variable, x, and we are treating y and z as constants, the "constant of integration" may be a function of y and z.)

    Since [itex]u= x^2/2+ f(y,z)[/itex], [itex]\partial u/\partial y= \partial f/\partial y= 2y[/itex]. Integrating with respect to y, [itex]f= y^2+ g(z)[/itex]. (We know that f is a function of y and z only. Since the derivative is with respect to the single variable, y, and we are treating z as a constant, the "constant of integration" may be a function of z.)

    Since [itex]f(x,y)= y^2+ g(z)[/itex], [itex]u(x,y,z)= x^2/2+ f(y,z)= x^2/2+ y^2+ g(z)[/itex] and so [itex]\partial u/\partial z= dg/dz= 3z[/itex]. (Notice that since g is a function of the single variable, z, this is an ordinary derivative.) Integrating [itex]g(z)= 3z^2/2+ C[/itex]. (Since g was a function of z only, this "constant of integration really is a constant!)

    Since [itex]u(x, y, z)= x^2/2+ y^2+ g(z)[/itex] we now have [itex]u(x, y, z)= x^2/2+ y^2+ 3z^2/2+ C[/itex].
  8. Oct 15, 2013 #7
    HallsofIvy, thank you. That was precisely what I was looking for. Although, I have one question. In post number three (one of my posts), is going from integral (1) to integral (2) incorrect? If so, why? If is it correct, how can one justify the move?
  9. Oct 16, 2013 #8


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    Your formula does not specify the variables other than the integration variable. How would you apply it for case b?

    You need to integral F along the chosen path. If it is (0,0,0)-->(x,0,0)-->(x,y,0)-->(x,y,z)
    [tex]\Delta U = - \int_{0,0,0}^{x,0,0} F_x (x',0,0) dx' - \int_{x,0,0}^{x,y,0} F_y(x,y',0) dy' - \int_{x,y,0}^{x,y,z´} F_z (x,y,z') dz'[/tex]

    In case a, Fx=Fx(x) Fy=Fy(y), Fz=F(z), all integrands contained only the integration variable, so your integral gave the correct result.

    In case b, Fx=ky, Fy=kx. If you integral Fx with respect to x' from 0 to x, you need the value of y along the integration path. What is it?

    For the path (0,0,0)-->(x,0,0)-->(x,y,0)-->(x,y,z), as before,

    [tex]\Delta U = - \int_{0,0,0}^{x,0,0}( 0) dx' - \int_{x,0,0}^{x,y,0} (kx) dy' - \int_{x,y,0}^{x,y,z´}(0) dz'=-kxy +C[/tex].

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